Stochastic Process (ENPC) Monday, 22nd of January 2018 (2h30)

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1 Stochastic Process (NPC) Monday, 22nd of January 208 (2h30) Vocabulary (english/français) : distribution distribution, loi ; positive strictement positif ; 0,) 0,. We write N Z,+ and N N {0}. We use the convention inf. xercice (Distribution of the maximum of a random walk). Let X be a Z-valued random variable and (X n, n N ) be independent random variables distributed as X. We assume that P(X 2) 0, p P(X ) > 0, X < + and m X < 0. We consider the random walk S (S n, n N) defined by S 0 0 and S n+ S n + X n+ for n N, and its natural filtration F (F n,n N). Our aim is to study the distribution of the global maximum of S : I Preliminaries. Prove that P(M < + ). M sups n. n N 2. Check that S is a Markov chain. Is it irreducible, transient, recurrent? 3. Check that the first hitting time of the level k N for the random walk S : τ k inf{n 0; S n k} is a stopping time with respect to filtration F. 4. Is the random time η inf{n 0; S n M} a stopping time? For λ 0, we set : ϕ(λ) e λx. 5. Prove that ϕ is of class C over (0,+ ), and check it is strictly convex. Prove there exists a unique λ 0 > 0 such that ϕ(λ 0 ). Check that ϕ (λ 0 ) > 0. II An auxiliary Markov chain We set q k e λ 0k P(X n k) for k Z, so that according to the previous question q (q k,k Z) is a probability distribution. Let (Y n,n N ) beindependent Z-valued random variables with probability distribution q. We introduce the random walk V (V n, n N) defined by V 0 0 and V n+ V n +Y n+ for n N.. Prove that for all measurable bounded or non-negative function f, we have : f(y,...,y n )e λ 0V n f(x,...,x n ). 2. Check that Y is integrable and that Y > 0. Let G (G n,n N) be the natural filtration of V and ρ k inf{n 0; V n k} be the first hitting time of the level k N for the random walk V. 3. Let k N. Check that V (k) (V n ρk,n N) is a sub-martingale (with respect to the filtration G) bounded from above by k. 4. Prove that V (k) a.s. converges. Deduce that ρ k is a.s. finite and compute V ρk.

2 5. Prove that ( e λ 0V n,n ) is a martingale (with respect to the filtration G). Compute : ρk ne λ 0V n G ρk n. 6. Deduce that P(τ k n) e λ 0k P(ρ k n), and then compute P(τ k < ). 7. Prove that M has a shifted geometric distribution, that is P(M k) α 0 ( α 0 ) k for k N. Write α 0 using λ 0. III The simple random walk Weconsiderthesimplerandomwalk withnegative drift:p(x ) p andp (0,/2).. Prove that P(S n < 0; n ) ( p)p(m 0). 2. Compute α 0 and deduce that the probability, P(S n < 0; n ), for the simple random walk S with negative drift to never come back to {0} is equal to 2p. xercice 2 (SomeBrownian martingale). Let B (B t,t R + ) beastandard Brownian motion and F (F t,t R + ) its natural filtration. For t 0,), we set : X t t e B2 t /2( t). We recall that the density of the standard Gaussian distribution N(0,) is given by : f(x) e x2 /2, x R, and that if G is a standard Gaussian random variable then, for all λ C, we have : e λg e λ2 /2.. Let 0 t s+t <. Compute X t+s F t. 2. Deduce that (X t,t 0,) is a continuous martingale. And deduce X t for all t 0,). 3. Prove that a.s. lim t X t Deduce that sup X t +. t 0,) 2

3 Correction I Preliminaries xercice. By the law of large number we have that a.s. lim n + S n /n m < 0. This implies that a.s. lim n + S n and thus that a.s. M < The process S is a stochastic dynamical system and thus a Markov chain. Since P(X ) > 0 and since there exists k N such that P(X k) > 0 (as m < 0), we deduce that S is irreducible. Since lim n + S n, we get that S is transient. 3. We have {τ k > n} {S j < k, 0 j n} F n for all n N and thus τ k is a stopping time with respect to F. 4. The random time η is not a stopping time because {η 0} n N {S n < 0} doesn t belong to F Notice that X n e λx max(e λ,sup x 0 x n e λx ), so the functions ϕ n (λ) X n e λx, for λ (0,+ ), are well defined for n N and locally bounded. Using Fubini, we get that b a ϕ n+(r)dr ϕ n (b) ϕ n (a) for all 0 < a b < +. This implies that ϕ is of class C over (0,+ ) with n-th derivative ϕ n. Since X 2 e λx is non-negative and positive with positive probability, we deduce that ϕ 2 > 0 and thus ϕ is strictly convex on (0, ). By dominated convergence, we get that ϕ and ϕ are continuous over 0,+ ), and we have ϕ(0), ϕ(+ ) +, ϕ (0) m < 0. The strict convexity of ϕ implies the existence of a unique root of ϕ(λ) on (0,+ ), say λ 0. Furthermore, we have that ϕ (λ 0 ) > 0. II An auxiliary Markov chain. We have that : {Y k,...,y nk n}e λ 0V n n j q kj e λ 0 n j k j This proves the result as the random variables X l and Y l are discrete. n p kj {X k,...,x nk n}. 2. We deduce from the previous question that Y X e λ 0X e λ 0 X < +. Thus Y is integrable and we have Y Xe λ 0X ϕ (λ 0 ) > Clearly V n is integrable, G n -measurable, and since Y n+ is independent of G n, we get Y n+ G n > 0. This implies that V is a sub-martingale. Similarly to Question I.3, we get that ρ k is a stopping time with respect to G. This gives that V (k) is the sub-martingale V stopped at the stopping time ρ k. Therefore, the process V (k) is a sub-martingale. By construction, we have V n < k on {n < ρ k }. Since Y n > 0 implies a.s. that Y n, we deduce that V ρk k on {n ρ k < }. In conclusion, V (k) is a sub-martingale bounded from above by k. 4. Since (V n ρk,n N) is a sub-martingale bounded from above by k, we deduce that it converges a.s. to a limit (which is integrable). Since the a.s. convergence of V (k) holds in Z, we get that either ρ k < or V (k) is constant for large time. The latter condition implies that the sequence (Y n,n N ) is constant for large n. This being impossible, we deduce that a.s. ρ k < +. Since ρ k is a.s. finite, we get that a.s. V ρk k. 5. Set N n e λ 0V n for n N. Notice that the process N (N n,n N) is G-adapted and non-negative. We also have : N n+ F n N n e λ 0Y n+ G n N n e λ 0Y n+ N n, j 3

4 where we used that N n is G n -measurable for the first equality, that Y n+ is independent of G n for the second and for the third that, by definition of the distribution of Y n+, we have e λ 0Y n+ e λ 0 X n+ +λ 0 X n+. We deduce that Nn+ N 0 and thus that N n+ is integrable. We have proven that N is a non-negative martingale. Notice that {ρ k n} belongs to G ρk n. By the stopping time theorem, since ρ k n is a bounded stopping time, we get : {ρk n}e λ 0V n G ρk n {ρk n} e λ 0V n F ρk n {ρk n}e λ 0V ρk n {ρk n}e λ 0k, where, for the last equality, we used that V ρk n V ρk k on {ρ k n}. 6. We have thanks to the previous question and Question II. that : P(τ k n) ρk ne λ 0V n ρk ne λ 0k e λ0k P(ρ k n). Letting n goes to infinity, and as ρ k is a.s. finite, we get that P(τ k n) e λ 0k. 7. Let k N. As {M k} {τ k < }, we deduce that P(M k) ( α 0 ) k with α 0 e λ 0. We deduce that P(M k) P(M k) P(M k +) α 0 ( α 0 ) k. III The simple random walk. Let M max n N S n X. Notice that M is distributed as M and is independent of X. Since {S n < 0; n } {X } {M 0}, we deduce that : P(S n < 0; n ) P(X )P(M 0) ( p)p(m 0). 2. We have ϕ(λ) pe λ +( p)e λ. We get that e λ 0 (0,) is a root of x ( p)x 2 x+p 0. This gives e λ 0 p/( p). We deduce that α 0 e λ 0 ( 2p)/( p). This gives : ( p)p(m 0) ( p)α 0 2p. xercice 2. The process X (X t,t 0,)) is F-adapted. Since X t+s is non-negative, we can compute its conditional expectation. Set η 2 t s. Using that B t+s B t is independent of F t and distributed as sg, with G a standard Gaussian random variable, we get that : X t+s F t η e (B t+s B t+b t) 2 /2η 2 Ft η e B2 t /2η2 H(B t ), with H(x) e (B t+s B t+x) 2 x 2 /2η 2 e ( sg+x) 2 x 2 /2η 2. 4

5 We get with z y t/η : We get : H(x) η t η t e ( sy+x) 2 x 2 /2η 2 y 2 /2 dy e y2 (s+η 2 )/2η 2 sxy/η 2 dy e z2 /2 sxz/η t dz e sxg/η t η t e sx2 /2η 2 ( t). η e x2 /2η 2 H(x) t e x2 /2( t). 2. We deduce from the previous question that X t+s F t X t. This gives, with t 0, that X s for all s 0,). Thus, as X t is non-negative, we have that X t L for all t 0,). Since the process X is F-adapted, we deduce that X is a martingale. Since B is continuous, we get that X is continuous. 3. As B 0 a.s. and B is continuous, we deduce there exists ε (0,) (random), such that B t 0 for all t ( ε,. Then use that lim x + xe x2 /2 0 to deduce that a.s. lim t X t Consider M (M n X /(n+),n N), which is a non-negative martingale with M 0 and a.s. lim n M n 0. If sup n N M n < +, then the martingale would also converge in L by dominated convergence. Since this not the case, we get that sup n N M n +. Then use that sup t 0,) X t sup n N M n to conclude. 5